Question

Let $M$ be a $3\times 3$ non-singular matrix with $det(M)=\alpha$. If $[M^{-1}adj(adj(M)]=KI$, then the value of $K$ is

• A. 1
• B. $\alpha$
• C. ${\alpha }^2$
• D. ${\alpha }^3$

Answer 1

Answer: (B)

Formulae:- $A^{-1}=\frac{1}{|A|}adj(A)$
$A\cdot adj(A)=|A|I=adj(A)\cdot A$
$adj(adjA)=|A{|}^{(n-2)}A$
Now, $[M^{-1}adj(adj(M)]$
$=\frac{1}{|M|}adj(M)\cdot |M{|}^{(3-2)}M$
$=adj(M)\cdot M$
$=|M|I$
$=\alpha I$
$\therefore k=\alpha$